Mixing
How much does a swing-by change the orbit of the planet?
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A friend of mine told me that one Voyager swing-by moved the orbit of Jupiter a foot closer to the sun. I understand that Jupiter was slowed by the gravity assist, but I figured it would be minuscule and basically unnoticeable. What is the true answer for this?
orbital-mechanics gravity-assist
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
$endgroup$
add a comment |
$begingroup$
A friend of mine told me that one Voyager swing-by moved the orbit of Jupiter a foot closer to the sun. I understand that Jupiter was slowed by the gravity assist, but I figured it would be minuscule and basically unnoticeable. What is the true answer for this?
orbital-mechanics gravity-assist
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
$endgroup$
$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
1
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@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
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– peterh
Jan 25 at 9:18
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Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
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– ben
Jan 27 at 1:22
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I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...]Its not a flyby, flyby does not use the object to increase speed necessarily
$endgroup$
– SinisterMJ
Jan 27 at 16:27
1
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41
add a comment |
$begingroup$
A friend of mine told me that one Voyager swing-by moved the orbit of Jupiter a foot closer to the sun. I understand that Jupiter was slowed by the gravity assist, but I figured it would be minuscule and basically unnoticeable. What is the true answer for this?
orbital-mechanics gravity-assist
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
$endgroup$
A friend of mine told me that one Voyager swing-by moved the orbit of Jupiter a foot closer to the sun. I understand that Jupiter was slowed by the gravity assist, but I figured it would be minuscule and basically unnoticeable. What is the true answer for this?
orbital-mechanics gravity-assist
orbital-mechanics gravity-assist
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
asked Jan 22 at 7:30
SinisterMJSinisterMJ
1213
1213
$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
1
$begingroup$
@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
$endgroup$
– peterh
Jan 25 at 9:18
$begingroup$
Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
$endgroup$
– ben
Jan 27 at 1:22
$begingroup$
I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...]Its not a flyby, flyby does not use the object to increase speed necessarily
$endgroup$
– SinisterMJ
Jan 27 at 16:27
1
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41
add a comment |
$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
1
$begingroup$
@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
$endgroup$
– peterh
Jan 25 at 9:18
$begingroup$
Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
$endgroup$
– ben
Jan 27 at 1:22
$begingroup$
I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...]Its not a flyby, flyby does not use the object to increase speed necessarily
$endgroup$
– SinisterMJ
Jan 27 at 16:27
1
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41
$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
1
1
$begingroup$
@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
$endgroup$
– peterh
Jan 25 at 9:18
$begingroup$
@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
$endgroup$
– peterh
Jan 25 at 9:18
$begingroup$
Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
$endgroup$
– ben
Jan 27 at 1:22
$begingroup$
Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
$endgroup$
– ben
Jan 27 at 1:22
$begingroup$
I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:
In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...] Its not a flyby, flyby does not use the object to increase speed necessarily$endgroup$
– SinisterMJ
Jan 27 at 16:27
$begingroup$
I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:
In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...] Its not a flyby, flyby does not use the object to increase speed necessarily$endgroup$
– SinisterMJ
Jan 27 at 16:27
1
1
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41
add a comment |
2 Answers
2
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oldest
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$begingroup$
- The mass of the Voyager is $approx$ 825 kg.
- The mass of the Jupiter is $approx 1.9 cdot 10^{27}$ kg.
According this question, the Voyager accelerated from $approx 10.2 frac{km}{s}$ to $approx 27.8 frac{km}{s}$ by its Jupiter flyby at 1979.7.29:
Thus, the Voyager got $approx 17.6 frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1
Step1
We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:
$$Delta v_{Voyager} cdot m_{Voyager} = Delta v_{Jupiter} cdot m_{Jupiter}$$
$$Delta v_{Jupiter} = frac{Delta v_{Voyager} cdot m_{Voyager}}{ m_{Jupiter} }$$
Substituting the values, we get $Delta v_{Jupiter} = 7.642 cdot 10^{-24} frac{km}{s}$.
Step2
Kepler's Third Law says:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)
Formalized:
$$P^2 propto a^3$$
Thus
$$P propto a^frac{3}{2}$$
Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).
According this source, the mean orbital velocity of the Jupiter is 13.1 $frac{km}{s}$ and its mean orbital radius is $7.786cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is
$$frac{7.64cdot10^{-24}}{13.1}cdot1.5cdot 7.786cdot10^8 km approx 6.811 cdot 10^{-16} km = underline{underline{6.811 cdot 10^{-13} m}}$$
The size of a nucleus is $approx 10^{-15}$ m. A foot is $approx 0.3 m$.
1Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $frac{km}{s}$ which is not a significant deviation.
answered Jan 22 at 7:47
peterhpeterh
2,03511531
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1
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So, a foot then @peterh!
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– GdD
Jan 22 at 9:09
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@GdD It is billionths of the size of an atom.
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– peterh
Jan 22 at 9:11
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I know @peterh, I was being sarcastic. Hard to communicate in a comment.
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– GdD
Jan 22 at 9:14
1
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If Jupiter were that massive, it would be a red dwarf.
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– Mark Adler
Jan 23 at 4:49
1
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Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
|
show 15 more comments
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I get that Voyager 1 brought Jupiter $6times10^{-13},mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.
Then Voyager 2 did almost that again! Oh, the humanity.
The $825,mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11,mathrm{km/s}$ by Jupiter, whose mass is $1.90times 10^{27},mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5times 10^{-21},mathrm{m/s}$.
Jupiter's velocity around the Sun at that time was $12.8,mathrm{km/s}$. So it was slowed down by $4times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2,mathrm{AU}$, you get $5.8times10^{-13},mathrm{m}$.
I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:
So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:
$$-{muover 2a}={v^2over 2}-{muover r}$$
The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:
$${muover 2a^2}da=v dv$$
We can rewrite that a little as:
$${muover 2a}{daover a}=v^2{dvover v}$$
If the orbit is circular, which Jupiter is close enough to, then:
$$v=sqrt{muover a}$$
So the $v^2$ and $mu/a$ cancel giving:
$${daover a}=2{dvover v}$$
Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
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I like this answer most because you worked backwards from loss of velocity.
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– Magic Octopus Urn
Jan 23 at 17:46
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Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
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– BlueCoder
Jan 24 at 11:18
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I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
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– BlueCoder
Jan 24 at 11:19
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Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
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– SinisterMJ
Jan 24 at 13:49
1
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Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
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– Mark Adler
Jan 25 at 16:33
|
show 2 more comments
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2 Answers
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2 Answers
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$begingroup$
- The mass of the Voyager is $approx$ 825 kg.
- The mass of the Jupiter is $approx 1.9 cdot 10^{27}$ kg.
According this question, the Voyager accelerated from $approx 10.2 frac{km}{s}$ to $approx 27.8 frac{km}{s}$ by its Jupiter flyby at 1979.7.29:
Thus, the Voyager got $approx 17.6 frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1
Step1
We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:
$$Delta v_{Voyager} cdot m_{Voyager} = Delta v_{Jupiter} cdot m_{Jupiter}$$
$$Delta v_{Jupiter} = frac{Delta v_{Voyager} cdot m_{Voyager}}{ m_{Jupiter} }$$
Substituting the values, we get $Delta v_{Jupiter} = 7.642 cdot 10^{-24} frac{km}{s}$.
Step2
Kepler's Third Law says:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)
Formalized:
$$P^2 propto a^3$$
Thus
$$P propto a^frac{3}{2}$$
Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).
According this source, the mean orbital velocity of the Jupiter is 13.1 $frac{km}{s}$ and its mean orbital radius is $7.786cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is
$$frac{7.64cdot10^{-24}}{13.1}cdot1.5cdot 7.786cdot10^8 km approx 6.811 cdot 10^{-16} km = underline{underline{6.811 cdot 10^{-13} m}}$$
The size of a nucleus is $approx 10^{-15}$ m. A foot is $approx 0.3 m$.
1Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $frac{km}{s}$ which is not a significant deviation.
answered Jan 22 at 7:47
peterhpeterh
2,03511531
$endgroup$
1
$begingroup$
So, a foot then @peterh!
$endgroup$
– GdD
Jan 22 at 9:09
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@GdD It is billionths of the size of an atom.
$endgroup$
– peterh
Jan 22 at 9:11
$begingroup$
I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
1
$begingroup$
If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
1
$begingroup$
Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
|
show 15 more comments
$begingroup$
- The mass of the Voyager is $approx$ 825 kg.
- The mass of the Jupiter is $approx 1.9 cdot 10^{27}$ kg.
According this question, the Voyager accelerated from $approx 10.2 frac{km}{s}$ to $approx 27.8 frac{km}{s}$ by its Jupiter flyby at 1979.7.29:
Thus, the Voyager got $approx 17.6 frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1
Step1
We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:
$$Delta v_{Voyager} cdot m_{Voyager} = Delta v_{Jupiter} cdot m_{Jupiter}$$
$$Delta v_{Jupiter} = frac{Delta v_{Voyager} cdot m_{Voyager}}{ m_{Jupiter} }$$
Substituting the values, we get $Delta v_{Jupiter} = 7.642 cdot 10^{-24} frac{km}{s}$.
Step2
Kepler's Third Law says:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)
Formalized:
$$P^2 propto a^3$$
Thus
$$P propto a^frac{3}{2}$$
Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).
According this source, the mean orbital velocity of the Jupiter is 13.1 $frac{km}{s}$ and its mean orbital radius is $7.786cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is
$$frac{7.64cdot10^{-24}}{13.1}cdot1.5cdot 7.786cdot10^8 km approx 6.811 cdot 10^{-16} km = underline{underline{6.811 cdot 10^{-13} m}}$$
The size of a nucleus is $approx 10^{-15}$ m. A foot is $approx 0.3 m$.
1Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $frac{km}{s}$ which is not a significant deviation.
answered Jan 22 at 7:47
peterhpeterh
2,03511531
$endgroup$
1
$begingroup$
So, a foot then @peterh!
$endgroup$
– GdD
Jan 22 at 9:09
$begingroup$
@GdD It is billionths of the size of an atom.
$endgroup$
– peterh
Jan 22 at 9:11
$begingroup$
I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
1
$begingroup$
If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
1
$begingroup$
Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
|
show 15 more comments
$begingroup$
- The mass of the Voyager is $approx$ 825 kg.
- The mass of the Jupiter is $approx 1.9 cdot 10^{27}$ kg.
According this question, the Voyager accelerated from $approx 10.2 frac{km}{s}$ to $approx 27.8 frac{km}{s}$ by its Jupiter flyby at 1979.7.29:
Thus, the Voyager got $approx 17.6 frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1
Step1
We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:
$$Delta v_{Voyager} cdot m_{Voyager} = Delta v_{Jupiter} cdot m_{Jupiter}$$
$$Delta v_{Jupiter} = frac{Delta v_{Voyager} cdot m_{Voyager}}{ m_{Jupiter} }$$
Substituting the values, we get $Delta v_{Jupiter} = 7.642 cdot 10^{-24} frac{km}{s}$.
Step2
Kepler's Third Law says:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)
Formalized:
$$P^2 propto a^3$$
Thus
$$P propto a^frac{3}{2}$$
Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).
According this source, the mean orbital velocity of the Jupiter is 13.1 $frac{km}{s}$ and its mean orbital radius is $7.786cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is
$$frac{7.64cdot10^{-24}}{13.1}cdot1.5cdot 7.786cdot10^8 km approx 6.811 cdot 10^{-16} km = underline{underline{6.811 cdot 10^{-13} m}}$$
The size of a nucleus is $approx 10^{-15}$ m. A foot is $approx 0.3 m$.
1Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $frac{km}{s}$ which is not a significant deviation.
answered Jan 22 at 7:47
peterhpeterh
2,03511531
$endgroup$
- The mass of the Voyager is $approx$ 825 kg.
- The mass of the Jupiter is $approx 1.9 cdot 10^{27}$ kg.
According this question, the Voyager accelerated from $approx 10.2 frac{km}{s}$ to $approx 27.8 frac{km}{s}$ by its Jupiter flyby at 1979.7.29:
Thus, the Voyager got $approx 17.6 frac{km}{s}$ velocity from the Jupiter by its gravitational slingshot maneuver.1
Step1
We use the Law of the Conservation of Momentum to calculate the change of the velocity of the Jupiter:
$$Delta v_{Voyager} cdot m_{Voyager} = Delta v_{Jupiter} cdot m_{Jupiter}$$
$$Delta v_{Jupiter} = frac{Delta v_{Voyager} cdot m_{Voyager}}{ m_{Jupiter} }$$
Substituting the values, we get $Delta v_{Jupiter} = 7.642 cdot 10^{-24} frac{km}{s}$.
Step2
Kepler's Third Law says:
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit. (ref)
Formalized:
$$P^2 propto a^3$$
Thus
$$P propto a^frac{3}{2}$$
Thus, a little change in the orbital velocity of the Jupiter causes the change of its 1.5 times in its orbital radius (more exactly, in the major axis of its elliptical orbit, but in the case of such little changes we can neglige this nuance).
According this source, the mean orbital velocity of the Jupiter is 13.1 $frac{km}{s}$ and its mean orbital radius is $7.786cdot10^8 km$. Thus, the change in the semi-major axis of the Jupiter's elliptical orbit is
$$frac{7.64cdot10^{-24}}{13.1}cdot1.5cdot 7.786cdot10^8 km approx 6.811 cdot 10^{-16} km = underline{underline{6.811 cdot 10^{-13} m}}$$
The size of a nucleus is $approx 10^{-15}$ m. A foot is $approx 0.3 m$.
1Mark Adler mentions in comment, that this picture is not a very good one to estimate the speed gain. However, my this source says 16 $frac{km}{s}$ which is not a significant deviation.
answered Jan 22 at 7:47
peterhpeterh
2,03511531
edited Feb 3 at 0:23
answered Jan 22 at 7:47
peterhpeterh
2,03511531
answered Jan 22 at 7:47
peterhpeterh
2,03511531
answered Jan 22 at 7:47
peterhpeterh
2,03511531
2,03511531
1
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So, a foot then @peterh!
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– GdD
Jan 22 at 9:09
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@GdD It is billionths of the size of an atom.
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– peterh
Jan 22 at 9:11
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I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
1
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If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
1
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Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
|
show 15 more comments
1
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So, a foot then @peterh!
$endgroup$
– GdD
Jan 22 at 9:09
$begingroup$
@GdD It is billionths of the size of an atom.
$endgroup$
– peterh
Jan 22 at 9:11
$begingroup$
I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
1
$begingroup$
If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
1
$begingroup$
Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
1
1
$begingroup$
So, a foot then @peterh!
$endgroup$
– GdD
Jan 22 at 9:09
$begingroup$
So, a foot then @peterh!
$endgroup$
– GdD
Jan 22 at 9:09
$begingroup$
@GdD It is billionths of the size of an atom.
$endgroup$
– peterh
Jan 22 at 9:11
$begingroup$
@GdD It is billionths of the size of an atom.
$endgroup$
– peterh
Jan 22 at 9:11
$begingroup$
I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
$begingroup$
I know @peterh, I was being sarcastic. Hard to communicate in a comment.
$endgroup$
– GdD
Jan 22 at 9:14
1
1
$begingroup$
If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
$begingroup$
If Jupiter were that massive, it would be a red dwarf.
$endgroup$
– Mark Adler
Jan 23 at 4:49
1
1
$begingroup$
Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
$begingroup$
Also the velocity change of Voyager is a) based on a very poor plot of Voyager 2, and more significantly, b) the velocity change is not read from the top of that spike. The top of the spike is the maximum velocity at closest approach, which is then mostly lost on departure, with the momentum returned to Jupiter. You need to read the velocity change from the points before and after the swingby where the Sun dominates the trajectory.
$endgroup$
– Mark Adler
Jan 27 at 17:00
|
show 15 more comments
$begingroup$
I get that Voyager 1 brought Jupiter $6times10^{-13},mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.
Then Voyager 2 did almost that again! Oh, the humanity.
The $825,mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11,mathrm{km/s}$ by Jupiter, whose mass is $1.90times 10^{27},mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5times 10^{-21},mathrm{m/s}$.
Jupiter's velocity around the Sun at that time was $12.8,mathrm{km/s}$. So it was slowed down by $4times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2,mathrm{AU}$, you get $5.8times10^{-13},mathrm{m}$.
I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:
So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:
$$-{muover 2a}={v^2over 2}-{muover r}$$
The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:
$${muover 2a^2}da=v dv$$
We can rewrite that a little as:
$${muover 2a}{daover a}=v^2{dvover v}$$
If the orbit is circular, which Jupiter is close enough to, then:
$$v=sqrt{muover a}$$
So the $v^2$ and $mu/a$ cancel giving:
$${daover a}=2{dvover v}$$
Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
$endgroup$
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
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– BlueCoder
Jan 24 at 11:18
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I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
1
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
|
show 2 more comments
$begingroup$
I get that Voyager 1 brought Jupiter $6times10^{-13},mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.
Then Voyager 2 did almost that again! Oh, the humanity.
The $825,mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11,mathrm{km/s}$ by Jupiter, whose mass is $1.90times 10^{27},mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5times 10^{-21},mathrm{m/s}$.
Jupiter's velocity around the Sun at that time was $12.8,mathrm{km/s}$. So it was slowed down by $4times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2,mathrm{AU}$, you get $5.8times10^{-13},mathrm{m}$.
I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:
So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:
$$-{muover 2a}={v^2over 2}-{muover r}$$
The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:
$${muover 2a^2}da=v dv$$
We can rewrite that a little as:
$${muover 2a}{daover a}=v^2{dvover v}$$
If the orbit is circular, which Jupiter is close enough to, then:
$$v=sqrt{muover a}$$
So the $v^2$ and $mu/a$ cancel giving:
$${daover a}=2{dvover v}$$
Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
$endgroup$
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
$endgroup$
– BlueCoder
Jan 24 at 11:18
$begingroup$
I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
1
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
|
show 2 more comments
$begingroup$
I get that Voyager 1 brought Jupiter $6times10^{-13},mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.
Then Voyager 2 did almost that again! Oh, the humanity.
The $825,mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11,mathrm{km/s}$ by Jupiter, whose mass is $1.90times 10^{27},mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5times 10^{-21},mathrm{m/s}$.
Jupiter's velocity around the Sun at that time was $12.8,mathrm{km/s}$. So it was slowed down by $4times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2,mathrm{AU}$, you get $5.8times10^{-13},mathrm{m}$.
I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:
So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:
$$-{muover 2a}={v^2over 2}-{muover r}$$
The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:
$${muover 2a^2}da=v dv$$
We can rewrite that a little as:
$${muover 2a}{daover a}=v^2{dvover v}$$
If the orbit is circular, which Jupiter is close enough to, then:
$$v=sqrt{muover a}$$
So the $v^2$ and $mu/a$ cancel giving:
$${daover a}=2{dvover v}$$
Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
$endgroup$
I get that Voyager 1 brought Jupiter $6times10^{-13},mathrm{m}$ closer to the Sun. One hundredth the size of a hydrogen atom.
Then Voyager 2 did almost that again! Oh, the humanity.
The $825,mathrm{kg}$ (wet mass) Voyager 1 spacecraft was sped up $11,mathrm{km/s}$ by Jupiter, whose mass is $1.90times 10^{27},mathrm{kg}$. The speed up was calculated using the heliocentric specific energy of Voyager 1 before and after the swingby. By the conservation of momentum, Jupiter was then slowed down by about $5times 10^{-21},mathrm{m/s}$.
Jupiter's velocity around the Sun at that time was $12.8,mathrm{km/s}$. So it was slowed down by $4times 10^{-25}$ of its velocity. By the relation of orbit velocity to radius, Jupiter's semi-major axis would have been reduced by that fraction times two. Multiplying by $5.2,mathrm{AU}$, you get $5.8times10^{-13},mathrm{m}$.
I fixed Voyagers' velocity change from a previous version of this answer. I should have looked at my own plot here. Here is that same plot for Voyager 1:
So why two times the fractional change in velocity? The relation between the semi-major axis $a$ and the velocity of an orbit $v$ at some point is the familiar equation:
$$-{muover 2a}={v^2over 2}-{muover r}$$
The velocity change occurs in a short period of time over which $r$ can be considered to be a constant. We can then take the derivative:
$${muover 2a^2}da=v dv$$
We can rewrite that a little as:
$${muover 2a}{daover a}=v^2{dvover v}$$
If the orbit is circular, which Jupiter is close enough to, then:
$$v=sqrt{muover a}$$
So the $v^2$ and $mu/a$ cancel giving:
$${daover a}=2{dvover v}$$
Therefore the resulting small fractional change in the semi-major axis is twice the small fractional change in velocity.
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
edited Jan 28 at 1:57
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
answered Jan 23 at 5:07
Mark AdlerMark Adler
49.9k3128212
49.9k3128212
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
$endgroup$
– BlueCoder
Jan 24 at 11:18
$begingroup$
I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
1
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
|
show 2 more comments
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
$endgroup$
– BlueCoder
Jan 24 at 11:18
$begingroup$
I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
1
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
I like this answer most because you worked backwards from loss of velocity.
$endgroup$
– Magic Octopus Urn
Jan 23 at 17:46
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
$endgroup$
– BlueCoder
Jan 24 at 11:18
$begingroup$
Isn't there some conflict between the two answers? In this answer, Voyager 1 (722Kg) gets 280m/s, Jupiter loses 10-22 m/s and this means ending up 4x10-15 closer. In peterh answer, Voyager 2 (825Kg) gets 17600m/s (!), Jupiter loses 10-23 m/s (gave more speed to Voyager 2 by losing less?) and this meant ending up 10-20 closer (lost one magnitude less and this means a smaller movement by 5 orders of mangitude - is this correct too?).
$endgroup$
– BlueCoder
Jan 24 at 11:18
$begingroup$
I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
I've now read your comment in the other answer, the first inconsistency is probably due to Jupiter having 2 orders of magnitude more mass there (red dwarf!) :)
$endgroup$
– BlueCoder
Jan 24 at 11:19
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
$begingroup$
Marked this answer as correct, as the weight on Google says 10 to the power of 27, not 29.
$endgroup$
– SinisterMJ
Jan 24 at 13:49
1
1
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
$begingroup$
Not as competent as you seem to think. I too hastily pulled some HORIZONS data without plotting it, and I had completely missed the flyby date. Fixed now.
$endgroup$
– Mark Adler
Jan 25 at 16:33
|
show 2 more comments
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$begingroup$
You've made me curious as to the mass of an object required for Voyager to shift it a full meter.
$endgroup$
– Magic Octopus Urn
Jan 24 at 2:06
1
$begingroup$
@MagicOctopusUrn $approx 10^{14}$ times more, so in the order of $10^{16}$ kg. It is the mass of an around 20km moon/asteroid, like Lysithea.
$endgroup$
– peterh
Jan 25 at 9:18
$begingroup$
Hey just curious where does the term "swing-by" come from? I think it's clear to most people on here what you are referring to, but I've pretty much exclusively heard the term "flyby."
$endgroup$
– ben
Jan 27 at 1:22
$begingroup$
I don't know, it has several names. In German we would usually say Swing-By. As for wikipedia:
In orbital mechanics and aerospace engineering, a gravitational slingshot, gravity assist maneuver, or swing-by [...]Its not a flyby, flyby does not use the object to increase speed necessarily$endgroup$
– SinisterMJ
Jan 27 at 16:27
1
$begingroup$
JPL routinely uses the term "swingby" (no hyphen) for gravity assists. I'm not sure when that started. At least back in the early 70's. Possibly before.
$endgroup$
– Mark Adler
Jan 27 at 16:41